(a) Remember each component of the average
velocity vector is simply given by the x or y displacement over
change in time: Vx average = (xb - xa)/(tb
- ta) and
Vy average = (yb - ya)/(tb
- ta). For the case at hand: Vx average = (xb
- xa)/(tb - ta) = -3.8 m/s,
where you are given( tb - ta) = 12.0
seconds. Find (xb - xa) and follow the same
procedure for the y component. Find (yb - ya)
in the same way.

(b) Compute the magnitude of the displacement
vector whose x and y components are given by
(xb - xa) and (yb - ya).
Use the Pythagorean Theorem. Note the displacement vector =
avearge

3. (NEW
ED) READ EXAMPLE 3.1 . Great
problem for you future or current web designers out there using
the same procedures as previous problem. The position vector = 2.5t2*i
+ 5.0*t*j = x(t)*i + y(t)* j, unit vectors in bold.
(a) Remember each component of the
average velocity vector is simply given by the x or y displacement
over change in time: Vx average = (xb - xa)/(tb
- ta) and
Vy average = (yb - ya)/(tb
- ta). Here is an example for the
x-component of the problem:
x((t) = 2.5t2.
We evaluate x(2) and x(0) by plugging in the time values into the
formula for x(t). Compute x(2) - x(0), then compute [x(2) -
x(0)]/(2 - 0). Also compute: [y(2) - y(0)]/(2 - 0).
Once you get each component of the average velocity, find the
magnitude using the Pythagorean Theorem. Find direction by
determining the angle the vector makes with the x axis and the
quadrant it points. See Ch. 1 and 4 problems and icq's/ quiz
1.
(b) For each formula x(t) and y(t), compute the derivative: for
example dx/dt = 5.0*t. Differentiate and plug in the numbers for
each component at the 3 different times.
(c) Sketch qualitatively the curve of motion on an x-y axes like
we did with projectile motion and other examples. Hint: The
curve is a parabola since x is proportional to the square of
y.

MORE DISCUSSIONS TO FOLLOW;
WITH RESPECT TO PROJECTILE MOTION, THESE PROBLEMS
CAN BE CONSIDERED "WARM UP "SINCE THEY ARE EASIER BUT
ARE POWERFUL TEACHING AND LEARNING TOOLS.

10. (NEW ED) Think two dimensionally. Review the logic
beyond the take home virtual lab: click here.
Read book examples. In this case it is only matter of finding
the time from y motion equations and plugging that time into
an x-motion equation.

18. (NEW ED) SEE class notes and textbook
examples on this asymmetric path with vertex at the highest
point. Contrast this with previous exercise, where the
particle returns to the same vertical level. In this
problem, the particle lands at a lower level. General
comments: Remember Vy = Voy - gt, where Voy
= Vo*sin 51.0. You are given the time of flight. Vx is
constant. The acceleration is constant and points downward.

19. (NEW ED) General comments. There is a
nice derivation of the parabolic path on page 79. See
equation 3.27. You could use this expression to find the height
but please understand how to derive it. Of course you must use
other means to find the y-component of velocity since
equation 3.27 does not give that.

29. (NEW
ED) SEE LECTURE CLASS
EXAMPLES WHERE I SHOWED A PARTICLE IN UNIFORM CIRCULAR MOTION AT 4
OR SO POINTS ALONG THE PATH. SEE ALSO EXAMPLES 3.11 AND
3.12. Remember, the centripetal acceleration vector always
points to the center of the circle whereas the velocity
vector is always tangent to path.

For relative motion
problems, covered in the next 4 exercises, always
remember
VPAx = VPBx + VBAx
VPAy = VPBy + VBAy,
where we have written the x and y components of the
following equation
vector-VPA = vector-VPB + vector-VBA
. This simply states that the velocity of a particle P
relative to frame A equals the velocity of the particle relative
to frame B plus the velocity of frame B relative to frame A.
Remember the BART train examples. Let Frame B be the train. Let
Frame A be the station, and let P be the person on a skateboard
coasting along the floor inside the train.
Recall in one example I said if VPBx = 10 mph and VBAx
= 80 mph (Bart train's top speed), then VPAx = 90 mph
as expected.

32. (NEW ED) You are to find VPBx
in three cases when VBAx =13.0 m/s . Discussion of (a)
opens the door to the other parts:
(a) VPAx = VPBx + VBAx
translates to 18 .0 m/s = VPBx + 13.0 m/s. Solve
for VPBx .
(b) Same equation but now VPAx= -3.0 m/s.
(c) Use the same method to reach an obvious conclusion.

33. USE (NEW
ED)
VPAx = VPBx + VBAx
VPAy = VPBy + VBAy, .

A = EARTH FRAME

B = RIVER FRAME

P = BOAT

ON YOUR PAPER , LET UP BE NORTH (N), DOWN BE
SOUTH (S), RIGHT BE EAST (E) AND LEFT BE WEST (W). RIGHT
IS THE POSITIVE X DIRECTION AND UP IS THE POSITIVE Y
DIRECTION ON YOUR AXES.

VPAx = +0.40*cos45 (m/s)
VPAy = -0.40*sin45 (m/s)

Note: SOUTHEAST means 45 degrees south of
east, in the 4th quadrant; NORTHWEST means 45 degrees north of
west, in the second quadrant, etc.

VBAx = 0.50 m/s and VBAy = 0.

USING THIS INFO, FIND THE MAGNITUDE OF
VECTOR-VPB BY FINDING BOTH OF ITS COMPONENTS AND
APPLYING THE PYTHAGOREAN THEOREM TO THEM.
Find the direction by accounting for the signs of the components
which will give you the quadrant in which VECTOR-VPB
points. Find the related angle theta thIS vector makes with
the x-axis by evaluating tan theta = |VPBy | / | VPBx
| .

For problems 35 and 36, see
in-class lecture notes and book examples 3.14 and 3.15
for reference. With regard to #35, see nearly identical
lecture example I did with a river flowing to the right and a boat
velocity relative to the river directed upward on the
paper. Thus the boat velocity relative to the shore
was not straight upward on the paper.

MORE DISCUSSIONS SHOULD FOLLOW
BUT TRY THE REMAINING PROBLEMS USING THE ABOVE
CONCEPTS. THEY REPRESENT AN IMPORTANT FOUNDATION, LIKE "WARM
UPS" FOR THE MAIN RACE. FREE TO EMAIL ME !

56. (NEW ED)This problem is like the egg
and the professor, #80, Ch. 2. The only difference is the
non-vertical launch direction toward the front of the incoming
boat. The horizontal displacement of the equipment
from the launch point is x = Vo*cosao*t.
Time t of course is obtained using Ch. 2 methods on
the y- motion equations for projectile motion. Where V
is the boat's speed, add V*t to the value of x to get the initial
distance D of the boat from shore,

62. (NEW ED)

(a) Go to page 84, figure 3.26 to see why you must always aim
above a target to hit it for any launch angle.
If you aim at or below the target you will never hit target
because of the downward pull of the gravitational force also known
as the weight. Note: The difference between the straight line and
the actual path is (1/2)*g*t2, simply seen by
evaluating

Vo*sinao*t - (Vo*sinao*t
- (1/2)*gt2) ,

the vertical difference between the straight line (no
gravity path) and actual trajectory.

If you want , rotate the figure shown and pretend you have a
horizontal launch to better see you must always aim above
target.

In any event find the angle if you aimed directly at the
target. To experiment with this idea find the angle then plug in
an angle less than this value into equation 3.27. For
the given y and x, there is no value of Vo
satisfying the equation.
(b) Use equation 3.27 but know how to derive it. Note 45 degrees
is larger than the angle of part (a). x
= Vo*sinao*t